The target that grows as the traveller slows

The circle light cannot leave found that a beam of light aimed at a black hole is swallowed if it passes within an impact parameter of $\sqrt{27}\,GM/c^2$, and that the disc of sky it removes, the shadow, is six and three-quarter times the cross-section of the horizon itself. The orbit that cannot be made smaller found the innermost stable circular orbit at $6GM/c^2$ and, inside it, a marginally bound orbit at $4GM/c^2$ which a particle falling from rest can circle but never settle on. The surface that only lets things in established what the horizon is.

None of those answers the question an astronomer asks first about any compact object moving through a crowd: how much of what drifts past does it swallow? The question has an answer for light, and the natural guess is that the answer for matter is about the same — that a black hole is a disc of a few horizon radii, and anything aimed inside it falls in. The guess is wrong by a factor that grows without limit as the matter slows down. This essay computes the capture cross-section for a particle of any speed and finds that it is set almost entirely by one number, the angular momentum $4GM/c$ per unit mass, and that for slow traffic a black hole is indistinguishable from an ordinary ball of four times its own radius.

The hill a slow particle has to climb #

A particle moving in the equatorial plane of a non-rotating hole has two conserved quantities, its energy and its angular momentum, both per unit rest mass. In units where $G = c = 1$ and lengths are measured in $M$, the radial motion obeys an energy equation with an effective potential

$$V(r) = \left(1 - \frac{2M}{r}\right)\left(1 + \frac{L^2}{r^2}\right),$$

and the particle can reach a radius only where its energy squared, $E^2$, exceeds $V$. A particle arriving from far away with speed $v$ has $E = \gamma = 1/\sqrt{1 - v^2}$, and its angular momentum is set by its aim: if it would pass at distance $b$ from the hole in the absence of gravity, then $L = \gamma v b$.

The figure takes the slowest possible arrival, $E^2 = 1$, a particle that falls in from rest at infinity. The potential has a bump for any angular momentum above $\sqrt{12}\,M$, and the bump grows as $L$ grows. At $L = 3.7\,M$ its peak is at 0.931, below the particle’s energy, so the particle rolls over it and into the hole. At $L = 4.3\,M$ the peak is at 1.080 and the particle is turned back. The dividing value is exactly $L = 4M$, where the peak reaches 1 at a radius of exactly $4M$ — the marginally bound orbit that the orbit that cannot be made smaller met from the other side.

The two numbers are the same because they are the same condition. The peak of the potential is a circular orbit — unstable, because it sits on a maximum — and a particle whose energy equals the peak’s height can arrive there and circle indefinitely. A slow particle has the energy of one at rest far away, so the circular orbit it can reach is the one whose energy is exactly that: the marginally bound one, at $4M$ with $L = 4M$.

Why aim matters so much when speed is small #

Everything else follows from $L = \gamma v b$. The capture condition is a condition on angular momentum, $L < L_c$, and a given angular momentum corresponds to an impact parameter $b = L/\gamma v$. For a slow particle $L_c = 4M$, so the critical impact parameter is

$$b_c \approx \frac{4GM}{c\,v},$$

and the cross-section is $\pi b_c^2 = 16\pi (GM/c^2)^2 (c/v)^2$.

The physical content is in the $1/v$. A particle aimed a long way off to one side carries angular momentum in proportion to its speed, so a slow one aimed far away carries very little. It is pulled inward, gains speed as it falls, and arrives near the hole with an angular momentum that was always small — too small, if its aim was inside $4GM/cv$, to hold up a centrifugal barrier against gravity’s strongest region. A fast particle aimed the same distance off carries a large angular momentum and is merely deflected.

The half of the bend a slow body never feels found a similar dependence on speed in the deflection of passing bodies; capture is its extreme. This is the same logic as gravitational focusing in Newtonian mechanics, and it is worth being precise about how the two differ, because the difference is the whole of the relativistic content.

The capture Newton cannot make #

In Newton’s theory the effective potential for a particle with angular momentum $L$ is $-GM/r + L^2/2r^2$. Whatever the value of $L$, the second term grows faster than the first as $r$ shrinks, and at some small radius it wins. A point mass therefore captures nothing: a particle with any angular momentum at all swings round the point and leaves. The only way for a Newtonian body to capture is to have a surface. A sphere of radius $R$ captures every particle whose orbit reaches $r < R$, and energy and angular momentum conservation give the familiar focused cross-section

$$\sigma_N = \pi R^2 \left(1 + \frac{2GM}{R\,v^2}\right).$$

For slow particles the second term dominates, and $\sigma_N \approx 2\pi GMR/v^2$ — growing as $1/v^2$ exactly as the black hole’s does. This is why a planet’s gravity makes it a far bigger target for slow debris than its disc would suggest, and why the slowest meteoroids dominate what it sweeps up.

Relativity changes the effective potential by one term. Expanding $V(r)$ gives, besides the Newtonian terms, $-2ML^2/r^3$, a correction that falls off faster than the centrifugal barrier at large distances but grows faster than it at small ones. At a few gravitational radii it overtakes the barrier and turns it over, and there is no longer a wall at small radius for the particle to bounce off. That is the capture. It needs no surface, because the potential itself falls away.

A target that grows as the traveller slows #

The exact cross-section at any speed is found by the same argument: for each speed, find the angular momentum at which the peak of $V$ equals $\gamma^2$, and convert it to an impact parameter. The figure plots the result as a multiple of the horizon’s own cross-sectional area, $\pi(2GM/c^2)^2$. At the speed of light it is $27/4 = 6.75$, the shadow of the circle light cannot leave. As the speed falls the curve bends up and joins the dashed line $4c^2/v^2$: 404 horizon areas at a tenth of light speed, forty thousand at a hundredth.

At the speed of the Galaxy’s dark matter, about 220 kilometres a second or $7.3 \times 10^{-4}$ of the speed of light, it is 7.4 million. For the four-million-solar-mass hole at the centre of the Galaxy, whose horizon radius is about $1.3 \times 10^{10}$ metres, that is a disc of radius about $3.5 \times 10^{13}$ metres — some 230 times the Earth’s distance from the Sun — within which a dark-matter particle’s path, left undisturbed by anything else, ends in the hole. The horizon itself is smaller than the orbit of Mercury.

The dotted curve is the Newtonian ball whose focused cross-section equals the black hole’s for slow particles. Matching $2\pi GMR/v^2$ to $16\pi(GM)^2/(c^2v^2)$ gives $R = 8GM/c^2$: four times the horizon radius, twice its diameter. For every particle slower than a few per cent of light speed the black hole and a solid ball of that size swallow the same things. The ball is not there, and the particles that it would have stopped at its surface are instead lost at an orbit half as far out.

The circle a marginal particle whirls on #

The critical case is worth following, because it shows where the capture actually happens.

At three-tenths of light speed the critical impact parameter is 13.88 gravitational radii. The four paths in the figure are aimed at 0.9 of it, at a millionth inside and a millionth outside, and at 1.12 times it. The first falls in after a single swing. The last is bent through a large angle and leaves. The two in the middle are indistinguishable for most of their approach: both arrive at the dashed circle, at 3.73 gravitational radii, and both circle the hole close to it. Then one drifts inward and crosses the horizon, and the other drifts outward and escapes.

That circle is the unstable orbit at the peak of the potential, and it is the real boundary. A particle is lost when it crosses it inward, not when it reaches the horizon, and whether it crosses was settled long before, by its aim. Near the critical value the number of turns a particle makes grows as the logarithm of how close its aim is. At three-tenths of light speed every factor of about 130 closer buys one more lap; for slow matter the factor is about 90, and for light it is $e^{2\pi}$, about 535. Some particles, therefore, circle many times before deciding, and a thin population of them spends a long time close to the hole.

The radius of that circle depends on the particle’s speed. The figure plots it: 4 gravitational radii for matter arriving with no speed, falling smoothly to 3 for light, the photon sphere, and passing 3.46 at half light speed. So the surface that decides capture is between one and a half and two horizon radii from the centre for every kind of traveller, and neither the horizon nor the photon sphere is it except for light. The last stable orbit, at 6, plays no part at all: capture is about unstable orbits, and the stable ones are where matter that has lost energy slowly — in a disc, say — ends its inspiral before plunging.

The ball a black hole pretends to be #

The Newtonian ball of radius $8GM/c^2$ matches the black hole only at low speeds. At higher speeds focusing matters less, and the equivalent ball must shrink to keep the same cross-section.

The solid curve is the radius of the Newtonian sphere that would capture particles of each speed at the rate the hole does. It falls from 8 for slow matter to 5.67 at half the speed of light and to 4.29 for light, which is where a Newtonian estimate for light — treating it as a particle focused by gravity onto a surface — would have to put the surface. The horizon, at 2, is well inside for every speed.

The dashed curve isolates the part of the answer that is genuinely relativistic. Plain focusing makes the critical impact parameter grow as $1/v$; the product $b_c v/c$ removes that and leaves a slowly varying factor which rises from 4 for slow particles to $\sqrt{27} = 5.20$ for light. So the whole of the speed-dependence across eight decades of kinetic energy is the $1/v^2$ of Newtonian focusing, multiplied by a number between 16 and 27. That is a compact statement worth carrying: the capture cross-section of a non-rotating hole is $\pi (GM/c^2)^2 (c/v)^2$ times a number that only climbs from 16 to 27 between a crawl and the speed of light.

What the number is used for #

A population of stars and compact objects around a galaxy’s central black hole is continually rearranged by encounters, and some of it is scattered onto orbits that carry it close to the hole. The theory of those rearrangements works in terms of a loss cone: the set of orbits whose angular momentum is small enough that they will be consumed on their next pass. For a compact object — a white dwarf, a neutron star, a stellar-mass black hole — orbiting a massive hole on a nearly parabolic orbit, the edge of the loss cone is exactly the critical angular momentum computed here, $4GM/c$ per unit mass. An object inside it plunges directly. An object just outside it can be caught by the emission of gravitational waves instead, losing energy on each pass and spiralling in over thousands of orbits — the slow inspirals that a space-based detector would observe in their tens of thousands of cycles. The line between the two fates is the one in the first figure.

Ordinary stars mostly do not get that far. For the Galaxy’s central hole a Sun-like star is torn apart by tides at about nine horizon radii, outside the capture orbit, so it is disrupted rather than swallowed whole, and the debris is gas rather than a test particle. For holes above about a hundred million solar masses the tidal radius falls inside the horizon, and a Sun-like star is swallowed whole with no flare at all. The capture cross-section here applies without correction to what survives tides: compact objects, and particles.

Dark matter is the purest case, because its particles are, as far as anyone knows, collisionless and too small to be disrupted. The seven-million-fold enhancement means that a black hole moving through a halo swallows dark matter at a rate far above the naive horizon-area estimate. The rate is still tiny in absolute terms, because dark matter is dilute: at a density of 0.4 GeV per cubic centimetre, the value near the Sun, a four-million-solar-mass hole would take in something like $10^{-11}$ solar masses a year by this route. What makes the capture matter is not the mass swallowed but the structure it carves. The same enhancement would govern much smaller holes: if small black holes were made in the early universe — the kind the hole that outlives everything and then does not followed to its end — the rate at which one sweeps up the Galaxy’s dark matter is set by this cross-section. A hole that sits at the bottom of a halo for billions of years empties the low-angular-momentum orbits around it, and how the remaining distribution rearranges itself is the subject of the question the last section leaves open.

Where the calculation stops #

The cross-section computed here assumes a non-rotating hole, a test particle with no size or charge, and a trajectory affected by nothing but the hole. Each assumption has a limit.

A spinning hole captures asymmetrically. A particle whose orbit runs in the same sense as the spin is dragged along with it and can approach more closely before being lost, while one moving against the spin is captured from further out, so the cross-section depends on the direction of approach and is larger on one side. The shadow of a spinning hole is displaced and flattened on one side for the same reason, and the figures here are the symmetric case that both reduce to at zero spin.

Collisions change everything. The calculation treats each particle alone. Gas does not behave like that: its particles collide, share angular momentum and radiate energy, and accretion of gas proceeds through the pressure-supported flows of Bondi accretion or through a disc, with rates set by the gas’s sound speed and its ability to lose angular momentum, not by the collisionless capture orbit. A black hole’s growth by gas has almost nothing to do with the numbers here, and its growth by stars and dark matter has everything to do with them.

The particle’s own wavelength has been ignored. For a wave whose wavelength is comparable with the horizon — no light from a star is, but gravitational waves from a merger are, and so would be the waves of a sufficiently light particle — the orbit picture fails and the capture must be computed as the absorption of a wave. For massless waves at very long wavelength the absorption cross-section tends not to the shadow’s $27\pi (GM/c^2)^2$ but to the horizon’s own area, $16\pi(GM/c^2)^2$, a result that turns out to hold for a wide class of black holes and is one of the more suggestive facts in the subject. The particle picture in this essay is the short-wavelength limit, which is every case of astrophysical interest except those.

What the pictures cannot show #

The figures show single paths in one plane and a cross-section that assumes particles arrive uniformly over a disc. A real population arrives with a spread of speeds, and the capture rate of the whole population is the cross-section averaged over the speed distribution; because it grows as $1/v^2$, the average is dominated by the slowest members, and a population with a long tail of slow particles is captured much faster than its typical speed suggests. The figures also show only the moment of capture, not its consequence: a particle that crosses the capture orbit inward reaches the horizon in a few gravitational times as measured from far away — though the two clocks disagree about the fall — and adds its energy, not its rest mass, to the hole. And the black disc in the orbit figure is drawn at the horizon’s circumferential radius; the geometry near it is not Euclidean, and the picture is a map of coordinates, not a photograph.

Still open: what a black hole does to the dark matter around it #

A black hole that grows slowly at the centre of a halo of collisionless particles should pull the surrounding distribution inward, because each particle’s orbit adjusts to the deepening well while conserving its adiabatic invariants. Calculations of that growth, first made in 1999, predict a steep spike in the dark-matter density close to the hole, rising as a power of radius near $-2.3$ down to within a few horizon radii, where capture empties it. If the spike exists and dark matter annihilates, the centre of the Galaxy would glow in gamma rays far more brightly than otherwise expected, and the absence of such a glow already constrains some candidate particles.

Whether the spike survives is disputed. Mergers of galaxies bring second black holes that scatter the dark matter out; the dense cluster of stars around the central hole heats the particles and softens the cusp; and the hole’s own history — whether it grew slowly in place or arrived already large — decides whether the adiabatic argument applies at all. Estimates of the present density near the Galaxy’s centre therefore span several orders of magnitude, and the capture orbit, which sets the inner edge of any spike, is the one piece of the calculation nobody doubts.

The habit worth carrying away is to ask which conserved quantity actually does the protecting. Speed does not keep a particle out of a black hole; angular momentum does, and a slow particle aimed a long way off carries almost none of it. The capture cross-section is the Newtonian $1/v^2$ of focusing multiplied by a factor between sixteen and twenty-seven, and the only reason light is harder to catch than a crawling particle is that it arrives with more angular momentum for the same aim.